CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g) In the reaction above, 0.5487 grams of calcium carbide are completely consumed to produce acetylene gas, C2H2. What volume (in mL) will this gas occupy if it is collected at 43 degrees Celsius and 0.926 atm pressure

239.7mL

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

The balanced chemical equation in this question is as follows:

CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g)

From the equation, 1 mole of CaC2 produces 1 mole of ethylene gas, C2H2.

Using mole = mass/molar mass

Molar mass of CaC2 = 40 + 12(2)

= 40 + 24

= 64g/mol

mole = 0.5487/64

mole = 0.00857mol of CaC2

Hence, 0.00857mol of CaC2 produced 0.00857mol of C2H2

Based on the information provided, n = 0.00857mol, T = 43°C = 43 + 273 = 316K, p = 0.926 atm

PV = nRT

V = nRT/P

V = 0.00857 × 0.0821 × 316/0.926

V = 0.222/0.926

V = 0.2397L

In mL, volume = 0.2397 × 1000

= 239.7mL

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